MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Surface energy of a liquid drop depends on its surface area, since surface tension is constant for the same liquid.Explanation:For a spherical drop, surface energy is given by $E = T \times A$.Here $T$ is surface tension and $A = 4\pi r^2$ is the surface area.Since the liquid is the same, $T$ remains constant, so $E \propto A$.Let the radius of one small drop be $r$ and the radius of the large drop be $R$.When 1000 small drops combine to form one large drop, the total volume is conserved.So, $1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$.Cancelling common terms gives $1000 r^3 = R^3$.Taking the cube root, we get $R = 10r$.Now, the ratio of surface energy of the large drop to that of one small drop is:$\frac{E_{\text{large}}}{E_{\text{small}}} = \frac{4\pi R^2}{4\pi r^2} = \frac{R^2}{r^2}$.Substituting $R = 10r$, we get $\frac{(10r)^2}{r^2} = 100$.Thus, the required ratio is $100 : 1$.Answer:The ratio of surface energy of 1 large drop to 1 small drop is $100 : 1$.Correct option: A.